Hard
Next Palindrome Using Same Digits — Python
Full explanation · Time O(n) · Space O(1)
# Time: O(n)
# Space: O(1)
class Solution(object):
def nextPalindrome(self, num):
"""
:type num: str
:rtype: str
"""
def next_permutation(nums, begin, end):
def reverse(nums, begin, end):
left, right = begin, end-1
while left < right:
nums[left], nums[right] = nums[right], nums[left]
left += 1
right -= 1
k, l = begin-1, begin
for i in reversed(xrange(begin, end-1)):
if nums[i] < nums[i+1]:
k = i
break
else:
reverse(nums, begin, end)
return False
for i in reversed(xrange(k+1, end)):
if nums[i] > nums[k]:
l = i
break
nums[k], nums[l] = nums[l], nums[k]
reverse(nums, k+1, end)
return True
nums = list(num)
if not next_permutation(nums, 0, len(nums)//2):
return ""
for i in xrange(len(nums)//2):
nums[-1-i] = nums[i]
return "".join(nums)